$(a)$ If $A \neq 0$, and if $a_{ii} = 0$, for all $1\leq i \leq n$, then $A^n = 0$.

$(b)$ If $A \neq I$ and if $a_{ii} = 1$ for all $1\leq i \leq n$, then $A$ is not diagonalizable.

$(c)$ If $A \neq 0$, then $A$ is invertible.

$(a)$ is true as eigenvalues of the upper triangular matrices are diagonal elements and here all the eigenvalues are $0$. Hence $x^n=0$ is the characteristic equation and by Cayley-Hamilton theorem $A^n=0$.

I think this should be tagged as homework. Also I presume you meant you have no idea about (b) rather than about (a) (which you seem to have answered) so I've edited it.
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Clive NewsteadSep 16 '12 at 17:21

2 Answers
2

For (b), consider the minimal polynomial of the matrix. Its characteristic polynomial is plainly $(x-1)^n$. So what must its minimal polynomial be if it is to be diagonalisable? Can its minimal polynomial be this?

Hint 1 (mouse-over to reveal):

A matrix is diagonalisable if and only if its minimal polynomial has distinct linear factors.

Hint 2:

The minimal polynomial of a matrix divides its characteristic polynomial.